If `B=int(1)/(e^(x)+1)dx=-f(x)+C`, where C is the constant of integration and `e^(f(0))=2`, then the value of `e^(f(-1))` is

Options

A. 4
B. `e+1`
C. `2e`
D. 0

Answer

To solve the given problem step by step, we will start from the integral expression provided and derive the necessary values to find \( e^{f(-1)} \).

### Step 1: Understand the integral expression
We are given:
\[ B = \int \frac{1}{e^x + 1} \, dx = -f(x) + C \]
This means that the integral \( B \) can be expressed in terms of \( f(x) \) and a constant \( C \).

### Step 2: Simplify the integral
We can rewrite the integral:
\[ B = \int \frac{1}{1 + e^x} \, dx \]
To solve this integral, we will use the substitution:
\[ t = 1 + e^x \implies e^x = t - 1 \quad \text{and} \quad dx = \frac{dt}{e^x} = \frac{dt}{t - 1} \]
Thus, the integral becomes:
\[ B = \int \frac{1}{t} \cdot \frac{dt}{t - 1} \]

### Step 3: Break down the integral
Now we can separate the fractions:
\[ \frac{1}{t(t - 1)} = \frac{1}{t} - \frac{1}{t - 1} \]
So, we can write:
\[ B = \int \left( \frac{1}{t} - \frac{1}{t - 1} \right) dt \]
This gives us:
\[ B = \ln |t| - \ln |t - 1| + C = \ln \left( \frac{t}{t - 1} \right) + C \]

### Step 4: Substitute back for \( t \)
Now, substituting back \( t = 1 + e^x \):
\[ B = \ln \left( \frac{1 + e^x}{e^x} \right) + C = \ln \left( \frac{1 + e^x}{e^x} \right) + C = \ln \left( \frac{1 + e^x}{e^x} \right) + C \]

### Step 5: Relate \( B \) to \( f(x) \)
From the original equation:
\[ B = -f(x) + C \]
We can set:
\[ -f(x) = \ln \left( \frac{1 + e^x}{e^x} \right) + C \]
Thus:
\[ f(x) = -\ln \left( \frac{1 + e^x}{e^x} \right) - C = \ln \left( \frac{e^x}{1 + e^x} \right) - C \]

### Step 6: Find \( f(0) \)
We know \( e^{f(0)} = 2 \):
\[ f(0) = \ln \left( \frac{e^0}{1 + e^0} \right) - C = \ln \left( \frac{1}{2} \right) - C \]
Thus:
\[ e^{f(0)} = e^{\ln \left( \frac{1}{2} \right) - C} = \frac{1}{2} e^{-C} \]
Setting this equal to 2 gives:
\[ \frac{1}{2} e^{-C} = 2 \implies e^{-C} = 4 \implies C = -\ln(4) \]

### Step 7: Find \( f(-1) \)
Now we need to find \( f(-1) \):
\[ f(-1) = \ln \left( \frac{e^{-1}}{1 + e^{-1}} \right) + \ln(4) \]
Calculating:
\[ f(-1) = \ln \left( \frac{\frac{1}{e}}{1 + \frac{1}{e}} \right) + \ln(4) = \ln \left( \frac{1}{e} \cdot \frac{e}{e + 1} \right) + \ln(4) = \ln \left( \frac{1}{e + 1} \right) + \ln(4) \]
Thus:
\[ f(-1) = \ln \left( \frac{4}{e + 1} \right) \]

### Step 8: Calculate \( e^{f(-1)} \)
Finally, we find:
\[ e^{f(-1)} = \frac{4}{e + 1} \]

### Final Answer
Thus, the value of \( e^{f(-1)} \) is:
\[ \boxed{\frac{4}{e + 1}} \]